MHT CET · Maths · Continuity and Differentiability
The value of \(\mathrm{c}\) for the function \(\mathrm{f}(x)=\log x\) on \([1, \mathrm{e}]\) if LMVT can be applied, is
- A \(e-2\)
- B \(e+1\)
- C \(e-1\)
- D e
Answer & Solution
Correct Answer
(C) \(e-1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{f}(x)=\log x \\
& \Rightarrow \mathrm{f}^{\prime}(x)=\frac{1}{x}
\end{aligned}\)
By Lagrange's Mean value theorem,
\(\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{e})-\mathrm{f}(1)}{\mathrm{e}-1} \\
& \Rightarrow \frac{1}{\mathrm{c}}=\frac{\log \mathrm{e}-\log 1}{\mathrm{e}-1} \\
& \Rightarrow \frac{1}{\mathrm{c}}=\frac{1}{\mathrm{e}-1} \\
& \Rightarrow \mathrm{c}=\mathrm{e}-1
\end{aligned}\)
& \mathrm{f}(x)=\log x \\
& \Rightarrow \mathrm{f}^{\prime}(x)=\frac{1}{x}
\end{aligned}\)
By Lagrange's Mean value theorem,
\(\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{e})-\mathrm{f}(1)}{\mathrm{e}-1} \\
& \Rightarrow \frac{1}{\mathrm{c}}=\frac{\log \mathrm{e}-\log 1}{\mathrm{e}-1} \\
& \Rightarrow \frac{1}{\mathrm{c}}=\frac{1}{\mathrm{e}-1} \\
& \Rightarrow \mathrm{c}=\mathrm{e}-1
\end{aligned}\)
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